(-x^2+4x-2)=(x^2+3x-8)

Solution for (-x^2+4x-2)=(x^2+3x-8) equation:

(-x^2+4x-2)=(x^2+3x-8)

We move all terms to the left:

(-x^2+4x-2)-((x^2+3x-8))=0

We get rid of parentheses

-x^2+4x-((x^2+3x-8))-2=0


We calculate terms in parentheses: -((x^2+3x-8)), so:

(x^2+3x-8)

We get rid of parentheses

x^2+3x-8

Back to the equation:

-(x^2+3x-8)


We add all the numbers together, and all the variables

-1x^2+4x-(x^2+3x-8)-2=0

We get rid of parentheses

-1x^2-x^2+4x-3x+8-2=0

We add all the numbers together, and all the variables

-2x^2+x+6=0

a = -2; b = 1; c = +6;
Δ = b2-4ac
Δ = 12-4·(-2)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-2}=\frac{6}{-4}
=-1+1/2
$